Two harmonic waves are represented in SI units by $y_1(x,t)=0.2\sin(x-3.0t)$ and $y_2(x,t)=0.2\sin(x-3.0t+\phi)$ (i) Write an expression for the sum $y=y_1+y_2$ for $\phi=\pi/2$ rad. (ii) Suppose the phase difference $\phi$ between the waves is unknown and the amplitude of their sum is 0.32 m,…

Question

Two harmonic waves are represented in SI units by $y_1(x,t)=0.2\sin(x-3.0t)$ and $y_2(x,t)=0.2\sin(x-3.0t+\phi)$ (i) Write an expression for the sum $y=y_1+y_2$ for $\phi=\pi/2$ rad. (ii) Suppose the phase difference $\phi$ between the waves is unknown and the amplitude of their sum is 0.32 m, what is $\phi$ ?

Aicharya · Accepted Answer

Sol. (i) $y=y_1+y_2=0.2\sin(x-3.0t)+0.2\sin\left(x-3.0t+\frac{\pi}{2}ight)$ $=A\sin(x-3.0t+	heta)$ Here, $A=\sqrt{(0.2)^2+(0.2)^2}=0.28\ \mathrm{m}$ and $	heta=\frac{\pi}{4}$ $	herefore\ y=0.28\sin\left(x-3.0t+\frac{\pi}{4}ight)$ (ii) Since, the amplitude of the resulting wave is 0.32 m and A = 0.2 m, we get $0.32=\sqrt{(0.2)^2+(0.2)^2+(2)(0.2)(0.2)\cos\phi}$ Solving this, we get $\phi=\pm 1.29\ \mathrm{rad}$ Answer: $\pm 1.29$ rad

Two harmonic waves are represented in SI units by $y_1(x,t)=0.2\sin(x-3.0t)$ and $y_2(x,t)=0.2\sin(x-3…

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